The question I had was
Find the Laplace transform of $$f(t)=10e^{-200t}u(t).$$
Would it be correct to take out the 10 because it is a constant, find the Laplace transform of $e^{-200t}$ and then multiply it by the Laplace transform of $u(t)$ to obtain a final answer of : $$10\left(\frac{1}{s+200}\right)\left(\frac{1}{s}\right)?$$ The $u(t)$ is what is really confusing me in this problem.
On
Another way to solve this problem is that the u(t) function means 1 for t>0,
0 for t <0 Since the laplace transform is restricted to the functions with t>0,i. e, for applying various properties of laplace transforms , functions must have t>0 condition, , So from earlier discussions our function becomes 10e^(-200t ), t>0 Now the laplace transform of 1 is 1/s and it is multiplied with 10e^(-200t ) so by shifting rule the laplace transform becomes 10/s+200. Just be careful while applying the properties or shortcut formulas of laplace transform for example if we have a function that is multiplied with Ua(t) i. e, u (t-a), then we can't directly just use the general shortcut formulas. Two methods can be used- 1. Either integrate from a to infinity.(laplace integral)
2 . or use the formula for u (t-a), Which is e^(-as)/s.
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Laplace Transform of $$e^{-200t}$$ does not exist
But Laplace transform of $$e^{-200t} u(t)$$ exist
$$ e^{-200t} u(t) \rightleftharpoons \frac{1}{s+200}$$
$$ 10e^{-200t} u(t) \rightleftharpoons \frac{10}{s+200}$$
And you can not multiply separately calculated Laplace transform
Multiplication in time domain gets converted into convolution in Laplace Domain
Yes, you can move the 10 out, but no, you cannot proceed as you do. Independent of your definition of Laplace transform (whole $\mathbb R$ or just $\mathbb R^+$ as domain of integration in the first step), you will find $$ (\mathcal L f)(s)=\int 10e^{-200t}u(t)e^{-st}\,dt=10\int_0^{+\infty}e^{-(200+s)t}\,dt=\frac{10}{s+200}. $$ As an alternative, you could also use rules for the Laplace transform: first that $\mathcal L u=1/s$ and that multiplication with the exponential shifts it, $1/(s+200)$ and that the multiplication of constant 10 just multiplies (since the Laplace transform is a linear operator), $10/(s+200)$.